Question

When a projectile is launched at an initial height of H feet above the ground at an angle of theta with the horizontal and initial velocity is Vo feet per second. the path of the projectile...

Answer 1

Given,

The initial height of H feet.

The initial velocity of the object is Vo.

The equation of the path of projectile is,

`y=h+x\text{ tan }\theta-(x^2)/(2V_0\cos ^2\theta)_{}\text{ }`

This is the expression of the projectle path.

Hence, the path of the projectile object is y = h + xtan(theta) - x²/2V₀²cos²(theta)